Verify that the left side is equal to the right side
$\displaystyle \frac{1+cos 3t}{sin3t} + \frac{sin 3 t}{1+ cos 3 t} = 2 csc 3 t$
$\displaystyle \frac{1 + \cos{3t}}{\sin{3t}} + \frac{\sin{3t}}{1 + \cos{3t}} = \frac{(1 + \cos{3t})^2}{\sin{3t}(1 + \cos{3t})} + \frac{\sin^2{3t}}{\sin{3t}(1 + \cos{3t})}$
$\displaystyle = \frac{1 + 2\cos{3t} + \cos^2{3t} + \sin^2{3t}}{\sin{3t}(1 + \cos{3t})}$
$\displaystyle = \frac{1 + 2\cos{3t} + 1}{\sin{3t}(1 + \cos{3t})}$
$\displaystyle = \frac{2 + 2\cos{3t}}{\sin{3t}(1 + \cos{3t})}$
$\displaystyle = \frac{2(1 + \cos{3t})}{\sin{3t}(1 + \cos{3t})}$
$\displaystyle = \frac{2}{\sin{3t}}$
$\displaystyle = 2\csc{3t}$.
$\displaystyle (\sec{u} - \tan{u})(\csc{u} + 1) = \sec{u}\csc{u} + \sec{u} - \csc{u}\tan{u} - \tan{u}$
$\displaystyle = \frac{1}{\cos{u}\sin{u}} + \frac{1}{\cos{u}} - \frac{\sin{u}}{\cos{u}\sin{u}} - \frac{\sin{u}}{\cos{u}}$
$\displaystyle = \frac{1 + \sin{u} - \sin{u} - \sin^2{u}}{\cos{u}\sin{u}}$
$\displaystyle = \frac{1 - \sin^2{u}}{\cos{u}\sin{u}}$
$\displaystyle = \frac{\cos^2{u}}{\cos{u}\sin{u}}$
$\displaystyle = \frac{\cos{u}}{\sin{u}}$
$\displaystyle = \cot{u}$.
$\displaystyle \frac{1}{\cos{u}\sin{u}} + \frac{1}{\cos{u}} - \frac{\sin{u}}{\cos{u}\sin{u}} - \frac{\sin{u}}{\cos{u}}$
$\displaystyle = \frac{1}{\cos{u}\sin{u}} + \frac{\sin{u}}{\cos{u}\sin{u}} - \frac{\sin{u}}{\cos{u}\sin{u}} - \frac{\sin^2{u}}{\cos{u}\sin{u}}$
$\displaystyle = \frac{1 + \sin{u} - \sin{u} - \sin^2{u}}{\cos{u}\sin{u}}
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